1121 Damn Single (25)

"Damn Single (单身狗)" is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party,so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case,the first line gives a positive integer N (<=50000),the total number of couples. Then N lines of the couples follow,each gives a couple of ID‘s which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples,there is a positive integer M (<=10000) followed by M ID‘s of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line,print their ID‘s in increasing order. The numbers must be separated by exactly 1 space,and there must be no extra space at the end of the line.

Sample Input:

3
22222 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 22222 23333

Sample Output:

题目大意：给出n对数，每一对数表示一对情侣的id；再给出m个id进行查询，找出没有情侣，或者情侣没有出现的人，按顺序输出这些人的id。
思路：1.用数组来记录情侣关系cp[i]=j,cp[j]=i 表示i,j是情侣关系，不存在情侣的记为cp[i]=-1。 
     2.记录每一个人的出现与否，exist[i]=1，表示i出现，exis[i]=0 表示i没有出现，将出现的人依次记录在present[i]中
　　　3.满足条件的为没有情侣的--cp[present[i]]=-1,以及情侣没有出现的--exist[cp[present[i]]]==0; 
要求按id顺序输出，用set来保存满足条件的id，插入过程中，自动排序，也可以保存在数组中，最后再排序
注意点：当 输入项比较多的时候，应该避免使用cin,cout这两个的效率较低5
10000 23333 44444 55555 88888

 1 #include<iostream>
 2 #include<vector>
 3 #include<set>
 4 using namespace std;
 5 int main(){
 6   vector<int> cp(100001,-1),present(100001),exist(100001,0);
 7   set<int> ans;
 8   int n,m,i,a,b,cnt=0;
 9   cin>>n;
10   for(i=0; i<n; i++){
11     scanf("%d%d",&a,&b);
12     cp[a] = b;
13     cp[b] = a;
14   }
15   cin>>m;
16   for(i=0; i<m; i++){
17     scanf("%d",&present[i]);
18     exist[present[i]] = 1;
19   }
20   for(i=0; i<m; i++){
21     if(!(cp[present[i]]!=-1 && exist[cp[present[i]]])){
22       cnt++;
23       ans.insert(present[i]);
24     }
25   }
26   cout<<cnt<<endl;
27   for(auto it=ans.begin(); it!=ans.end(); it++){
28     if(it==ans.begin())  printf("%05d",*it);
29     else  printf(" %05d",*it);
30   }
31 }